# X Incentivizing Participation in Online Forums for Education

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13 X:13 For x to be a best response, U(x, λ, µ) must be maximized at x. Note that U is concave in x, so setting the derivative to zero is a necessary and sufficient condition for an interior maximum. (n 1)λ + µ Taking the derivative and equating to zero, we have = α, which gives us (x + (n 1)λ + µ) (n 1)λ+µ x = α ((n 1)λ + µ). Since we seek a symmetric equilibrium, we want x = λ. This gives us (n 1)λ + µ = nλ + µ, α yielding the following quadratic equation in the variable λ: The solutions to this quadratic are, respectively, αn λ + λ(nµα (n 1)) + (αµ µ) = 0. (λ 1 (µ), λ (µ)) = n 1 αn µ n ± (n 1) 4α n 4 + µ αn 3. (n 1) Since 4α n + µ 4 αn n 1 3 αn, it follows that λ (µ) 0 for any µ. This cannot be the best response since λ is the rate of an exponential and hence must be positive. So the best response of the students when the instructor chooses a rate µ is either λ 1 (µ) if λ 1 (µ) 0, or else it is 0: λ(µ) = max{0, λ 1 (µ)} = max{0, n 1 αn µ (n 1) n + 4α n 4 + µ αn 3 }. Now we investigate the function λ 1 (µ). At µ = 0, λ 1 (0) = n 1 αn. The derivative of λ 1 with respect to µ tells us how the best response rate chosen by the students changes as the instructor increases her rate up from 0, if the best response is to participate: d dµ λ 1(µ) = 1 n αn 3 = (n 1) 4α n + µ 4 αn 3 1 n 1 1. αn (n 1) 4α n + µ 4 αn 3 d Note that dµ λ 1(µ) is decreasing in µ. At µ = 0, the derivative evaluates to d dµ λ 1(µ) = 1 [ ] 1 0 n n for all n. Since the derivative is decreasing in µ, and non-positive at µ = 0, this means that λ 1 (µ) is maximized at µ = 0, as long as n. Now, the value of λ 1 at µ = 0 is λ(0) = n 1 αn > 0, so the value of µ that maximizes λ(µ) = max{0, λ 1 (µ)} is µ = 0, as long as n. That is, if there are at least two students in the class (and we need at least two students for there to be a race between students to first supply the correct answer), the instructor can maximize the rate of response of the students by choosing the lowest possible rate for herself, i.e., by setting µ = 0. Note that for any n, λ 1 (µ) becomes 0 at µ 0 = 1 α. (Since the derivative of λ 1(µ) is decreasing in µ and non-positive at µ = 0, and λ 1 (0) > 0, λ 1 (µ) crosses zero exactly once for µ 0). Therefore, the students best response rate λ(µ) behaves as follows: λ(µ) = n 1 αn µ (n 1) n + 4α n 4 + µ αn 3, for µ [0, 1 α ] 0, for µ 1 α.

15 X:15 which goes to 0 as n diverges, so that the first term in U I (µ) also goes to 0 in n. Thus we see that for sufficiently large n, U I (µ) < 0 for µ < α 1. It follows that µ = 0 is a local maximum of the instructor s utility when n is large enough. Next we consider what happens if the instructor chooses a rate µ α 1. In this case the students will not participate so that the combined rate φ(µ) = µ, and so the instructor s utility is given by U I (µ) = µ βµ. If we consider this as a function over all µ 0, it is concave, with a unique µ+δ local maximum at µ = max { 0, } δ β δ. But this is the correct expression for the instructor s utility only if µ α 1, i.e., once the students have droppedout, so we need to ask when this local maximum µ exceeds α 1. The condition for µ > α 1 δ is β δ > 1 δ and hence β < α (δ + α 1. We thus have two cases (again keeping in ) mind that n is sufficiently large): δ When β < (δ + α 1 ), the instructor s utility has a second local maximum µ (in addition to µ = 0) at a value of µ beyond α 1 this local maximum corresponds to the instructor answering the question alone with no help from the class, at a rate which optimally trades off her cost to checking the forum and her benefit from the asker receiving an answer quickly. δ On the other hand, when β > (δ + α 1 ), then U I(µ) is monotonically decreasing as soon as µ α 1 this is because the concave function µ µ+δ βµ is decreasing beyond its maximum, i.e., for µ µ, and µ < α 1. Hence µ = 0 is the only local maximum for this range of β. δ Finally, in the case when β < (δ + α 1, we can ask which of the two local maxima 0 or µ ) achieves a higher utility. The utility at the local maximum µ = 0 is U I (0) = φ(µ) φ(µ) + δ = The utility at the other maximum µ is given by n 1 αn n 1 αn + δ αδ. U I (µ ) = (1 βδ). Now we observe that as β goes to 0, so that the instructor s effort becomes less and less costly, the utility U I (µ ) converges to 1 while U I (0) remains bounded below 1. Thus there is a β such that when β < β, the second local maximum, when the instructor answers the question on her own, has the higher utility. In the limit as n, the utility U I (0) 1 is converging to 1 + αδ, and so the cross-over point where U I(0) = U I (µ ) is converging to the solution of (1 βδ) 1 = 1 + αδ, which is β = 1 δ ( δα ).

16 X:16 5. MIXED FORUMS In this section, we address the question of how an instructor might elicit higher rates from students for single-answer questions, and show that one way to offset some of the slow behavior arising from single-answer questions is by ensuring that the forum contains a mixture of single-answer and open-ended questions. We imagine that each question is open-ended with probability p > 0 and single-answer with probability 1 p. Each student s λ parameter, however, is a global rate that governs how frequently he checks the site; by definition it must be chosen before any particular question is seen. When the student checks the site (according to rate λ), he may sometimes find an open-ended question, and sometimes a single-answer question. As before, we find that there is a unique symmetric equilibrium λ for any choice of µ. Our main result here is that the instructor can choose µ so as to elicit a rate λ that is roughly p times what it is in the case of purely open-ended questions, and this is tight for large n. In this way, the rate at which the 1 p fraction of single-answer questions get answered benefits significantly from the presence of even a small positive fraction p > 0 of open-ended questions they are essentially sped up by a linear amount (in n) when p is any constant. THEOREM 5.1. Let p be the probability that a question is open-ended. The instructor can choose a rate µ such that the equilibrium student rate λ satisfies λ p. Moreover this is essentially 4α tight in the limit, in the following sense: For any ε > 0, there exists a large enough n so that with at least n students, there is no µ the instructor can choose so as to achieve λ (1 + ε)p. 4α PROOF. Suppose the instructor chooses rate µ; we want to determine the best response rate λ chosen by the students. Using the payoffs derived in 3 and 4, the payoff to a student who chooses rate x when the instructor has rate µ and the remaining n 1 students choose rate λ is x x U(x) = p + (1 p) x + µ x + (n 1)λ + µ αx. As usual, for x to be a best response, a necessary condition is for U (x) to be zero, which gives us µ p (x + µ) + (1 p) (n 1)λ + µ (x + (n 1)λ + µ) α = 0. Since we are looking for a symmetric equilibrium, we set x = λ, which gives us the following equation in λ: µ 1)λ + µ p + (1 p)(n (λ + µ) (nλ + µ) = α. (4) Note that (4) can be interpreted as a bivariate equation in the two variables λ and µ; for a given value of µ, if there is a solution λ(µ) 0 to this equation, then λ(µ) is a best response to µ. p We first identify a choice of µ for which the equilibrium value λ(µ) is at least. It turns out 4α that this can be achieved using a µ for which λ(µ) = µ, as follows. In order to have an equilibrium in which λ(µ) = µ, we simply need a pair (µ, λ) satisfying Equation (4) for which µ = λ. Setting µ = λ in (4), we get leading to p λ nλ + (1 p) 4λ (n + 1) λ = α, λ = p + 4 n+1 (1 p) p 4α 4α, (5)

17 X:17 as desired. This gives us a lower bound on the fastest possible equilibrium λ, by demonstrating the value of λ achieved at a particular carefully-chosen equilibrium. Next we will show that as n becomes large, this lower bound is essentially the best possible. To do this, fix any constant ε > 0, choose σ small enough that (1 σ)(1+ε) > 1, and then choose n large enough in terms of ε, p, and σ so that 4 < σ. Now suppose by way of contradiction that pn for some choice of µ, there is a pair (µ, λ) that solves the bivariate equation (4) with λ Then we have µ 1)λ + µ α = p + (1 p)(n (λ + µ) (nλ + µ) µ nλ + µ p + (1 p) (λ + µ) (nλ + µ) µ = p (λ + µ) + (1 p) 1 (nλ + µ) µ p (λ + µ) + 1 nλ µ p (λ + µ) + 4α pn µ < p (λ + µ) + σα, (1 + ε)p. 4α and hence µ p > (1 σ)α. (λ + µ) p Writing r = (1 σ)α, this implies an upper bound on λ via the inequality λ < rµ µ. Now, over all µ 0, the right-hand side of this inequality is maximized at µ = r/4, where it takes the value r/4, and hence we have λ < r/4 = p 4(1 σ)α (1 + ε)p <. 4α (1 + ε)p This last inequality contradicts our assumption that λ. 4α Thus, as long as there is any positive probability p that a question in the forum will be an open-ended question (which might be discussion questions posed by the instructor herself), the instructor can choose µ to produce a rate λ such that the rate of arrival of the first response to any question on the forum scales linearly with the size of the class, obtaining a speedup with class size which is of the same order as in the case of purely open-ended questions. 6. DISCUSSION In this paper, we introduced a game-theoretic model for online forums for education, and investigated how an instructor can influence student participation and utility in these forums. For single-answer questions, we find that the instructor has very little influence over the rate at which students participate in the forum, especially as the class size grows larger and larger here, the rate of response is driven almost entirely by the students competition amongst each other to arrive first to supply the answer. However, when the instructor is optimizing for the speed at which questions are answered, taking into account her own ability to spend effort on producing answers, a

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